Is it a fact that under a continuous function, the preimage of an open, pathwise connected set is pathwise connected itself? I'm trying to prove that $GL_n(\mathbb{C})$ is pathwise connected, without explicitly constructing paths.

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    $\begingroup$ In general, no. Consider a projection $\pi \colon \mathbb{Z}\times X \to X$. $\endgroup$ Jul 23, 2013 at 14:51
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    $\begingroup$ Alright, thanks. Can you think of additional requirements that would turn it around? $\endgroup$ Jul 23, 2013 at 14:56
  • $\begingroup$ Path-connected fibres. But that would just reduce the problem to show that $SL_n(\mathbb{C})$ is path connected. $\endgroup$ Jul 23, 2013 at 15:00
  • $\begingroup$ Depending on what theory you can throw at it, you may be able to use the fact that the determinant is a nonconstant polynomial (hence holomorphic function), and the zero sets of those are too thin to separate the space. So if you have several complex variables or algebraic geometry to throw at it, you're done. Otherwise, rotate to make the first column a multiple of $e_1$, use the induction hypothesis to conclude. $\endgroup$ Jul 23, 2013 at 15:12
  • $\begingroup$ Complex analysis is a possibility, but what is "too thin" supposed to mean? I believe I would have to prove, that for any $M \in \mathbb{C}$ with $|M| < \mathbb{R}$, the space $\mathbb{C} \setminus M$ is pathwise connected. I can't think of a way to do that right now.. $\endgroup$ Jul 23, 2013 at 15:25

1 Answer 1


No, the preimage of a (pathwise) connected set under a continuous map has no reason to be connected. For examples, consider the projection $\pi \colon X\times Y \to X$ where $Y$ is a discrete space (with more than one point), or maps with discrete domain (these are automatically continuous).

I'm trying to prove that $GL_n(ℂ)$ is pathwise connected, without explicitly constructing paths.

Depending on what counts as explicit, we can prove it more or less easily.

First way, almost explicit paths:

Let $A \in GL_n(\mathbb{C})$. Consider $\varphi \colon \mathbb{C} \to M_n(\mathbb{C});\; \varphi(z) = (1-z)\cdot A + z\cdot I$. Then $p = \det \circ \varphi$ is a polynomial (of degree $\leqslant n^2$) with $p(0) \neq 0 \neq p(1)$, hence has only finitely many zeros. Choose a path $\gamma$ in $\mathbb{C}$ from $0$ to $1$ that avoids the zeros of $p$.

Then $\varphi \circ \gamma$ is a path in $GL_n(\mathbb{C})$ connecting $A$ and $I$.

That was easy.

Second way, no paths, but heavy machinery:

Definition: Let $D \subset \mathbb{C}^n$ open. A subset $E \subset D$ is thin in $D$ if it is locally contained in the zero set of a non-constant holomorphic function, i.e.

$$\bigl(\forall x\in D\bigr) \bigl(\exists U \in \mathcal{V}(x)\bigr) \bigl(\exists f \in \mathcal{O}(U)\setminus\{0\}\bigr)\bigl(E\cap U \subset Z(f)\bigr).$$

Proposition: Let $D \subset \mathbb{C}^n$ open and connected. Let $E \subset D$ thin in $D$. Then $D \setminus E$ is connected.

The proposition is proved in most (hopefully) books on several complex variables, for example, it's corollary 3.6 in chapter I of Range's "Holomorphic Functions and Integral Representations in Several Complex Variables".

Armed with that heavy machinery, we now observe that $E = M_n(\mathbb{C}) \setminus GL_n(\mathbb{C})$ is the zero set of a non-constant holomorphic function (the determinant), hence thin. Therefore $GL_n(\mathbb{C}) = M_n(\mathbb{C}) \setminus E$ is connected, since the vector space $M_n(\mathbb{C})$ is connected. $GL_n(\mathbb{C})$ is also open, hence it is pathwise connected.


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